Question Detail

There are two pipes which are functioning simultaneouly to fill a tank in 12 hours, if one pipe fills the tank 10 hours faster than other then how many hours second pipe will take to fill the tank ?

30 hours
35 hours
40 hours
42 hours

Answer: Option A

Explanation:

Lets suppose tank got filled by first pipe in X hours,
So, second pipe will fill the tank in X + 10 hours.

\begin{aligned}
=> \frac{1}{X} + \frac{1}{X} + 10 = \frac{1}{12} \\
=> X = 20
\end{aligned}

1. An electric pump can fill a tank in 3 hours. Because of a leak in the tank, it took 3 hours 30 min to fill the tank. In what time the leak can drain out all the water of the tank and will make tank empty ?

10 hours
13 hours
17 hours
21 hours

Answer: Option D

Explanation:

We can get the answer by subtrating work done by leak in one hour by subtraction of filling for 1 hour without leak and with leak, as

Work done for 1 hour without leak = 1/3
Work done with leak =
\begin{aligned}
3\frac{1}{2} = \frac{7}{2} \\
\text{Work done with leak in 1 hr= }\frac{2}{7} \\
\text{Work done by leak in 1 hr }\\
= \frac{1}{3} - \frac{2}{7} = \frac{1}{21}
\end{aligned}

So tank will be empty by the leak in 21 hours.

2. Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will the tank be filled ?

2.5 hours
2 hours
3.5 hours
3 hours

Answer: Option D

Explanation:

Part filled by A in 1 hour = 1/5
Part filled by B in 1 hour = 1/10
Part filled by C in 1 hour = 1/30

Part filled by (A+B+C) in 1 hour =
\begin{aligned}
\frac{1}{5}+\frac{1}{10}+\frac{1}{30} \\
= \frac{1}{3} \\
\end{aligned}

So all pipes will fill the tank in 3 hours.

3. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. Find the time required by the first pipe to fill the tank ?

10 hours
15 hours
17 hours
18 hours

Answer: Option B

Explanation:

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.

As per question, we get
\begin{aligned}
\frac{1}{x} + \frac{1}{x-5} = \frac{1}{x-9} \\
=> \frac{x-5+x}{x(x-5)} = \frac{1}{x-9}\\
=> (2x - 5)(x - 9) = x(x - 5)\\
=> x^2 - 18x + 45 = 0
\end{aligned}

After solving this euation, we get
(x-15)(x+3) = 0,
As value can not be negative, so x = 15

4. A cistern can be filled in 9 hours but due to a leak at its bottom it takes 10 hours. If the cistern is full, then the time that the leak will take to make it empty will be ?

20 hours
19 hours
90 hours
80 hours

Answer: Option C

Explanation:

Part filled without leak in 1 hour = 1/9
Part filled with leak in 1 hour = 1/10

Work done by leak in 1 hour \begin{aligned}
= \frac{1}{9} - \frac{1}{10} \\
= \frac{1}{90}
\end{aligned}

We used subtraction as it is getting empty.

So total time to empty the cistern is 90 hours

5. A tap can fill a tank in 6 hours. After half the tank is filled then 3 more similar taps are opened. What will be total time taken to fill the tank completely.

2 hours 30 mins
2 hours 45 mins
3 hours 30 mins
3 hours 45 mins

Answer: Option D

Explanation:

Half tank will be filled in 3 hours
Lets calculate remaining half,

Part filled by the four taps in 1 hour = 4*(1/6) = 2/3

Remaining part after 1/2 filled = 1-1/2 = 1/2

\begin{aligned}
\frac{2}{3}:\frac{1}{2}::1:X \\

=> X = \left( \frac{1}{2}*1*{3}{2} \right) \\
=> X = \frac{3}{4} hrs = 45 \text{ mins} \\

\end{aligned}
Total time = 3 hours + 45 mins = 3 hours 45 mins